Final answer:
To determine how many tons of CaTiO3 are needed to produce 5.00 tons of titanium metal, the limiting reactant concept must be used. Magnesium is the limiting reactant in the reaction, and the stoichiometry of the reaction shows that 1 mole of titanium is formed for every 2 moles of magnesium. Based on the moles of magnesium available, the mass of titanium can be calculated, and this value can be scaled up to find the corresponding mass of CaTiO3 required.
Step-by-step explanation:
To calculate how many tons of calcium titanate (CaTiO3) are required to produce 5.00 tons of titanium metal, we must first understand the process of titanium extraction and the limiting reactant concept.
The production of titanium involves the conversion of titanium tetrachloride (TiCl4) to metallic titanium using magnesium. This reaction is shown in Equation 8.5.2: TiCl4(g) + 2 Mg(l) → Ti(s) + 2 MgCl2(l). From the given information, 8.23 mol of magnesium reacts with 5.272 mol of titanium tetrachloride, but magnesium is the limiting reactant. Hence, the amount of titanium produced is based on the magnesium available.
From the stoichiometry of the reaction, it is clear that 1 mol of titanium is produced for every 2 mol of magnesium. With the molar mass of titanium being 47.867 g/mol, we can calculate the mass of titanium produced from the moles of magnesium available, which is 4.12 mol of Ti, corresponding to 197 g Ti. To scale this up to tons, consider that if 4.12 mol of Ti yields 0.197 kg, then 5.00 tons (5,000,000 g) would require a proportional amount of CaTiO3.